Question 515728: A rectangle is three times as long as it is wide. If its length and width are both decreased by 2 cm, its are is decreased by 36 sq. cm. Find its original dimensions.
I've been having trouble with this problem every time I try to solve it I get x=-40/12. I've checked the answer but my teacher wants us to do the process right not just get the right answer.
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! First, you must believe in your knowledge of the world, regardless of what the math says.
I have a formal logical proof that 1=2. That cannot be true, of course, but the proof looks good.
In this case, a rectangle cannot have negative length and width.
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L = length in cm
W = width in cm
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Length is three times width.
L = 3W
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If L and W are decreased by 2 cm
L-2
W-2
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Then its area is decreased by 36 cm^2
Original area is
A = L*W
so
(L-2)(W-2) = L*W -36
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substitute for L=3W
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(3W -2)(W-2) = 3W*W -36
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3W^2 -6W -2W +4 = 3W^2 -36
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3W^2 -8W + 4 -3W^2 = -36
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-8W + 4 = -36
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-8W = -40
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W = 5 cm is width
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L = 3W = 3(5) = 15 cm is the length
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Check to see of the areas compare correctly.
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L*W = 15*5 = 75 cm^2
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(L-2)*(W-2) = 13*3 = 39 cm^2
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75 - 39 = 36 cm^2
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Correct.
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Answer: The original dimensions are length = 15 cm and width = 5 cm.
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Done.
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